# Kinematics question projectiles with vector components (baseball player Q)

1. Mar 27, 2012

### chiuda

1. The problem statement, all variables and given/known data
Base ball player makes perfect contact with a ball striking it at 45° angle above the horizontal at a point 1.3 m above the ground. His ball just makes it clear of the 3m wall 130m from home plate. What was the velocity at with he struck the ball.

2. Relevant equations
is my final answer correct? i think my steps are good and that my assumptions are perfectly reasonable but i am getting a very unrealistic answer, please can somebody help me it would be very appreciated?! :)

3. The attempt at a solution

Ok so i assumed that when the question states the ball just makes it over the wall that its vertical velocity is 0m/s then assuming that i plugged it into this kinematic equation
d=(v2)(t)- (1/2)(a)(t)^2 (note i am trying to find time just through the vertical components) i then subtracted the height of which the ball was hit with where v2 was 0m/s which is 3m giving me 1.7 m for vert displacement so i have d i have v2 and i have a with is -9.8m/s^2. after plugging this into the equation and solving for t i get 0.6s. i then move on to find the horizontal velocity since it is constant and i now have time so i plug it into v=d/t because i have d horizontal (130m) and t (0.6s) from this i get 217 m/s (v horizontal) i then automatically know vi= 217 m/s as well because it was projected at a 45° angle then i use pathagreans theorem and solve for Vi and end up with 307 m/s [45°].

2. Mar 27, 2012

### tms

I would think it means that at 130 m the ball was 3 m above the ground. It's vertical speed won't be zero until it hits the ground (and bounces a few times, but that can be ignored).

3. Mar 27, 2012

### chiuda

yes i understand that but when the ball reaches its maximum height it has a vertical velocity which I'm assuming that the 3 m above ground is its maximum height

4. Mar 27, 2012

### tal444

3 m above ground is NOT the maximum height. That is only the height of the fence, the ball needs to be hit much higher if it is to clear the fence. And it should be obvious that the ball could not be travelling at over 1000km/h initially...

Hint: What you need to find out is the t, perhaps in a form involving another variable.

5. Mar 27, 2012

### chiuda

I know that it's unrealistic that's why I asked the question I know I need to find t but it's unknown and I don't have enough information to find it! Ahhhhhhhhhh!

6. Mar 27, 2012

### tal444

Like I said... t will end up containing another variable. That's okay, because it will cancel out in the end. t = $\frac{d}{v}$. What is your horizontal d and horizontal v?

7. Mar 27, 2012

### chiuda

i only have dh no vh or t my d horizontal is 130 m

8. Mar 27, 2012

### tal444

Again... Stop trying to plug in numbers for now. What is your horizontal velocity in terms of variables only?

9. Mar 27, 2012

### chiuda

horiz velocity is d

10. Mar 27, 2012

### chiuda

sorry horiz velocity is v

11. Mar 27, 2012

### chiuda

horiz displacement is d

12. Mar 27, 2012

### tal444

No... v is your velocity at 45°. What is your horizontal velocity, if the velocity at 45° is v?

Hint: Use the Pythagorean theorem.

13. Mar 27, 2012

### chiuda

Ok horizon velocity is Vh

14. Mar 27, 2012

### tal444

... NO. If my hypotenuse is v, then my horizontal component of my v must equal vcos45. That's what you need to plug in.

15. Mar 28, 2012

### chiuda

OK. I don't have the horizontal component of the initial velocity

16. Mar 28, 2012

### azizlwl

$x=u_xt$
$y=u_yt-\frac{1}{2}at^2$

$u_x=uCos45°$
$u_y=uSin45°$

At time t, the ball is 130 from base(x) and at height of 3m(y)
You have 2 equations,2 unknowns.

17. Mar 28, 2012

### chiuda

OMG I JUST DID IT WITH A FRIEND THANK YOU SO MUCH!!! Sorry for the confusion, but that one was a doozy